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Bull. Korean Math. Soc. 55 (2018), No. 4, pp. 1161–1178 https://doi.org/10.4134/BKMS.b170634

pISSN: 1015-8634 / eISSN: 2234-3016

GLOBAL SOLUTIONS FOR A CLASS OF NONLINEAR SIXTH-ORDER WAVE EQUATION

Ying Wang

Abstract. In this paper, we consider the Cauchy problem for a class of nonlinear sixth-order wave equation. The global existence and the finite time blow-up for the problem are proved by the potential well method at both low and critical initial energy levels. Furthermore, we present some sufficient conditions on initial data such that the weak solution exists globally at supercritical initial energy level by introducing a new stable set.

1. Introduction

This paper is concerned with the Cauchy problem for the following nonlinear sixth-order wave equation

utt− uxxtt− uxx+ uxxxx+ uxxxxtt= f (ux)x, (1.1)

u(x, 0) = φ(x), ut(x, 0) = ψ(x), (1.2)

where x ∈ R, u(x, t) is the unknown function, f (s) is a given nonlinear function.

In order to investigate the water wave problem with surface tension, Schnei- der and Wayne [12] considered a class of Boussinesq equation which models the water wave problem with surface tension as follows

utt= uxx+ uxxtt+ µuxxxx− uxxxxtt+ (u2)xx, (1.3)

where x, t, µ ∈ R and u(t, x) ∈ R. The model can also be formally derived from the 2D water wave problem. Eq. (1.3) with µ > 0 is known as the “good”

Boussinesq equation because of its linear instability. For a degenerate case, they have proved that the long wave limit can be described approximately by two decoupled Kawahara-equations. The authors included the term with the sixth-order derivatives since they were interested precisely in the case when the coefficient in front of the term with the fourth-order derivative is small, i.e., 1 + µ = ε2ν, with ν ∈ R fixed [12]. The lowest order nonlinear terms in the water wave problem remain unchanged from the classical equation because they

Received July 19, 2017; Revised December 2, 2017; Accepted December 29, 2017.

2010 Mathematics Subject Classification. 35L60, 35K55, 35Q80.

Key words and phrases. Cauchy problem, sixth-order wave equation, blow-up, existence of global solution.

c

2018 Korean Mathematical Society 1161

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are independent of surface tension. Wang [20, 21] studied the well-posedness of the local and global solution, the blow-up of solutions and nonlinear scattering for small amplitude solutions to Eq. (1.3) in R and Rn. The global existence and finite time blow-up of the solutions for Eq. (1.3) are established by the potential well method [17]. Eq. (1.1) is similar to the generalized Boussinesq equation and various generalized of the Boussinesq equations have been studied from many aspects [5, 6, 8–10, 13–15].

Wang and Xu [16] considered the Cauchy problem for the following Rosenau equation

utt+ uxxxx+ uxxxxtt− ruxx= f (u)xx. (1.4)

Under some conditions, the well-posedness of the solution and the nonexistence of global solution to the problem are proved with the aid of the potential well method. The authors [23] studied the following nonlinear wave equation

utt+ uxxxx+ uxxxxtt− ruxx= φ(ux)x. (1.5)

The existence and nonexistence of global solutions to Eq. (1.5) are obtained by the potential well method. Most of the authors proved the global existence and finite time blow-up of the solution at the sub-critical initial energy level (E(0) < d) and critical initial energy level (E(0) = d) by the potential well- method (the definitions of E(t) and d will be given later). In particular, to our knowledge, there have only been a few results up to now on the global existence of a solution to the Cauchy problem (1.1) and (1.2) at the high initial energy level (E(0) > 0).

Hatice et al. [18, 19] considered the following Rosenau equation (1.6) utt− uxx+ uxxxx+ uxxxxtt= (f (u))xx, x ∈ R, t > 0,

where f (u) = γ|u|p, γ > 0. By defining new functionals and using potential well method, they established the existence of global weak solutions for Eq. (1.6) with supercritical initial energy (E(0) > 0). Furthermore, the authors general- ized the results from one dimension to n-dimensional spaces. Then, Kutev et al. [4] considered the Cauchy problem for the Boussinesq paradigm equation (1.7) utt− ∆u − β1∆utt+ β22u = ∆f (u), (x, t) ∈ Rn× (0, +∞), where f (u) = α|u|p, α > 0, β1≥ 0, β2> 0, and ∆ denotes the Laplace operator in Rn. They introduced some new functionals and gave the existence of global weak solution with supercritical initial energy E(0) > 0. But the method in [4, 18, 19] is not valid for the nonlinear term f (u) = β|u|pu, β < 0 at arbitrarily positive initial energy level E(0) > 0.

Wang and Mu [21] considered the following Boussinesq equation (1.8) utt− ∆utt+ ∆2utt− ∆u + ∆2u = ∆f (u), (x, t) ∈ Rn× (0, +∞).

They obtained the existence and the uniqueness of the global solution and blow- up of the solution under some restrictions on the nonlinear source term f (u).

When f (u) = ±β|u|por −β|u|p−1u, β > 0, Xu et al. [24] considered the Cauchy

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problem of Eq. (1.8) at three different initial energy levels. They established the global existence and blow-up solutions at low and critical initial energy levels, and also proved the global existence of the weak solution at supercritical initial energy level. In this paper, we apply the potential well method ([7, 22, 24]) to study the global existence and nonexistence of the solution to the problem (1.1) and (1.2) at the sub-critical initial energy level, the critical initial energy level and the high initial energy level.

In the present paper, using the contraction mapping principle, the well- posedness for problem (1.1) and (1.2) was established in Section 2. Under some conditions of f (u), we also prove the existence and uniqueness of the global solution for the problem (1.1) and (1.2) in Section 2. In sections 3 and 4, we fix the nonlinear terms to be f (u) = r|u|p, r 6= 0 to classify the initial data for the global existence and non-global existence. Following the main idea of potential well method introduced in [7, 22, 24], we will construct the stable and unstable sets and study the global existence and non-existence of the weak solutions for problem (1.1) and (1.2) at three different initial energy levels:

E(0) < d, E(0) = d and E(0) > 0, where d is the potential well depth, and the three cases will be respectively dealt with different methods.

Throughout this paper, Lp denotes the usual space of all Lp(R)-functions with norm k · kp and kuk = kuk2, Hs denotes the Sobolev space Hs(R) with norm k·kHsand H01(R) denotes the closure of Cc(R) in H1(R). If not specified throughout this paper, we denote C as a generic constant varying line by line and depending only on the norms of initial data and absolute constant.

At first, by using the contraction mapping theorem, we obtain the following existence and uniqueness of the local solution to problem (1.1) and (1.2).

Theorem 1.1. Suppose that s > 12, φ, ψ ∈ Hs and f ∈ CN(R), then there exists a maximal time T0which depends on φ and ψ such that for each T < T0, the Cauchy problem (1.1) and (1.2) has a unique solution u ∈ C1([0, T ], Hs).

Moreover, if

sup

t∈[0,T0)

[ku(·, t)kHs+ kut(·, t)kHs] < ∞, then T0= ∞.

Theorem 1.2. Suppose that s ≥ 1, φ, ψ ∈ Hs, F (u) =Ru

0 f (z)dz, F (φx) ∈ L1 and T0 > 0 is the maximal existence time of corresponding solution u(t) ∈ C1([0, T0), Hs) to the Cauchy problem (1.1) and (1.2). Then the equality

E(t) = 1

2[kutk2+ kuxtk2+ kuxk2+ kuxxk2+ kuxxtk2] + Z

R

F (ux)dx

= E(0), ∀t ∈ (0, T0), (1.9)

holds.

Under some assumptions, the well-posedness of the global solution for the Cauchy problem (1.1) and (1.2) is established.

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Theorem 1.3. Suppose that the assumptions of Theorem 1.1 hold, and T0> 0 is the maximal existence time of the corresponding solution u(t) ∈ C2([0, T0], Hs) to the problem (1.1) and (1.2). Then T0< ∞ if and only if

lim

t→T0sup kux(t)kL = ∞.

For the case f (s) = r|u|p, r 6= 0, we firstly introduce the potential energy functional

(1.10) J (u) = 1

2(kuxk2+ kuxxk2) + r p + 1

Z

R

|ux|puxdx and the Nehari functional

(1.11) I(u) = (kuxk2+ kuxxk2) + r Z

R

|ux|puxdx.

We define the stable set

(1.12) K1= {u ∈ H2| I(u) > 0} ∪ {0}, the unstable set

(1.13) K2= {u ∈ H2| I(u) < 0}

and the depth of potential well as d = inf

u∈N EJ (u),

where the Nehari manifold N E = {u ∈ H2\ {0} | I(u) = 0}. And for the function u(x, t) satisfying u ∈ C1((0, T ), H2), ut ∈ C((0, T ), H2), we define a new functional space

YT := {u|I(u(t)) > kutk2+ kuxtk2+ kuxxtk2} ∪ {0}, which will be used in Section 4.

For the low initial energy case and the critical energy case, we prove the global existence and finite-time blow-up for the problem (1.1) and (1.2) by the potential well method.

Theorem 1.4. Suppose that 2 ≤ s < p + 1 and φ, ψ ∈ Hs. If E(0) ≤ d and φ ∈ K1∪ ∂K1, then problem (1.1) and (1.2) has a unique solution u ∈ C1([0, ∞); Hs) and u(t) ∈ K1∪ ∂K1 for t ∈ [0, ∞).

Theorem 1.5. Suppose that 2 ≤ s < p+1 and φ, ψ ∈ Hs. If E(0) < d, φ ∈ K2

and (φ, ψ) + (φx, ψx) + (φxx, ψxx) ≥ 0 when E(0) = d, then the solution u(x, t) of problem (1.1) and (1.2) ceases to exist in finite time.

For the supercritical initial energy case, by utilizing the method of [4,24], we obtain the global existence of weak solution for problem (1.1) and (1.2) under some sufficient conditions on the initial data.

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Theorem 1.6. Suppose that 2 ≤ s < p + 1 and φ ∈ H2∩ H01, ψ ∈ H2∩ H01. If E(0) > 0,

(1.14)

I(φ) > kψk2+ kψxk2+ kψxxk2, (1.15)

2(φ, ψ) + 2(φx, ψx) + 2(φxx, ψxx)

+ kφk2+ kφxk2+ kφxxk2+2(p + 1)

p + 3 E(0) < 0, (1.16)

then the solution u(x, t) of problem (1.1) and (1.2) exists globally.

2. Well-posedness of solutions

In this section, we study the local and global well-posedness of solutions to the problem (1.1) and (1.2). We first consider the following linear wave equation

utt− uxxtt− uxx+ uxxxx+ uxxxxtt= q(x, t), (2.1)

with initial value (1.2). By the Fourier transform and Duhamel’s principle, the solution u(x, t) of problem (2.1) and (1.2) can be written as

(2.2) u(x, t) = (∂tS(t)φ)(x) + (S(t)ψ)(x) + Z t

0

Γ(t − τ )q(u(τ ))dτ.

Here Γ(t) = S(t)(1 − ∂2x+ ∂x4)−1 and (∂tS(t)φ)(x) = 1

2π Z

R

eixξcos( |ξ|p 1 + ξ2t

p1 + ξ2+ ξ4) ˆφ(ξ))dξ, (S(t)ψ)(x) = 1

2π Z

R

eixξsin( |ξ|p 1 + ξ2t p1 + ξ2+ ξ4)

p1 + ξ2+ ξ4

|ξ|p 1 + ξ2

ψ(ξ))dξ,ˆ

where φ(ξ) = F (φ)(ξ) =ˆ R

Re−ixξφ(x)dx is the Fourier transform of φ(x).

Lemma 2.1. For the operators ∂tS(t), S(t) and Γ(t), we have the following estimates

k∂tS(t)φkHs ≤ kφkHs, ∀φ ∈ Hs, (2.3)

kS(t)ψkHs≤ 2(1 + t)kψkHs, ∀ψ ∈ Hs, (2.4)

k∂ttS(t)φkHs−2 ≤√

2kφkHs, ∀φ ∈ Hs, (2.5)

kΓ(t)qkHs ≤√

2kqkHs−4, ∀q ∈ Hs−4, (2.6)

k∂tΓ(t)qkHs−2 ≤ kqkHs−4, ∀q ∈ Hs−4. (2.7)

Proof. We only prove (2.4) and (2.6), the proof of other inequalities are similar.

By Plancherel’s theorem, we obtain kS(t)ψk2Hs =

Z

R

(1 + ξ2)s1 + ξ2+ ξ4

ξ2(1 + ξ2) sin2( t|ξ|p 1 + ξ2

p1 + ξ2+ ξ4)| ˆψ(ξ)|2

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≤ Z

|ξ|≤1

(1 + ξ2)st2| ˆψ|2dξ + Z

|ξ|>1

(1 + ξ2)s(1 + ξ2+ ξ4)

ξ2(1 + ξ2) | ˆψ(ξ)|2

≤ t2 Z

|ξ|≤1

(1 + ξ2)s| ˆψ(ξ)|2dξ + 4 Z

|ξ|>1

(1 + ξ2)s| ˆψ(ξ)|2

≤ 4(1 + t)2kψk2Hs,

kΓ(t)qk2Hs= Z

R

(1 + ξ2)ssin2(t|ξ|

1+ξ2

1+ξ24)1 + ξ2+ ξ4 ξ2(1 + ξ2)

1

(1 + ξ2+ ξ4)2|ˆq(ξ)|2

≤ 2 Z

R

(1 + ξ2)s−4|ˆq(ξ)|2dξ = 2kqk2Hs−4.

Therefore (2.4) and (2.6) hold. This completes the proof of the lemma.  Lemma 2.2 ([1]). Suppose that g(u) ∈ CN(R) is a function vanishing at zero, where N ≥ 0 is an integer. Then for any s with 0 ≤ s ≤ N and any u, v ∈ Hs∩ L, we have

kg(u)kHs ≤ G(kukL)kukHs,

kg(u) − g(v)kHs ≤ ¯G(kukL, kvkL)ku − vkHs,

where G : [0, ∞) → R and ¯G : [0, ∞) × [0, ∞) → R are continuous functions.

Lemma 2.3 (Sobolev Lemma [2]). If s > n2 + k, where k is a nonnegative integer, then

Hs(Rn) ⊂ Ck(Rn) ∩ L, where the inclusion is continuous. In fact,

Σ|β|≤kk∂βxukL ≤ CkukHs, where C is dependent on s, independent of u.

Lemma 2.4 ([11]). Assume u ∈ Hs∩ L, 0 < s < p. Then there exists a constant C such that

k|u|pkHs ≤ Ckukp−1LkukHs.

Lemma 2.5 ([3]). If s > 0, then Hs∩ L is an algebra. Moreover, kuvkHs ≤ C(kukLkvkHs+ kvkLkukHs).

Lemma 2.6. The operator P = (1 − ∂2x+ ∂x4)−1 is bounded from Hs−4to Hs, i.e.,

kP ukHs ≤ CkukHs−4. Proof.

kP ukHs = k(1 + ξ2)s2 1 1 + ξ2+ ξ4ukˆ

≤ Ck(1 + ξ2)s−42 uk = Ckukˆ Hs−4. 

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Proof of Theorem 1.1. Now we are going to prove the existence and uniqueness of local solutions for problem (1.1) and (1.2) by contraction mapping argumen- tation. For this purpose, we define the function space X(T ) = C1([0, T ]; Hs) with s > 12, equipped with the norm defined by

kukX(T )= max

t∈[0,T ]

[ku(·, t)kHs+ kut(·, t)kHs].

Since Hs,→ L for s > 12, we have u ∈ L if u ∈ X(T ). Let BR(T ) be the ball of radius R centered at the origin in X(T ), i.e.,

BR(T ) = {u ∈ X(T ) : kukX(T )≤ R}.

For φ ∈ Hs, ψ ∈ Hs, u ∈ X(T ), we define the map (2.8) Θ(u(t)) = ∂tS(t)φ + S(t)ψ +

Z t 0

S(t − τ )(I − ∂2x+ ∂x4)−1f (ux)x(τ )dτ.

It will be shown that Θ : BR(T ) → BR(T ) is contractive if R and T are well chosen.

Let kφkHs + kψkHs ≤ ρ, u, v ∈ BR(T ). Using the estimates in Lemmas 2.1-2.5, we have

kΘ(u)kHs ≤ kφkHs+ 2(1 + T )kψkHs−2+√ 2

Z T 0

kf (ux))kHs−3

≤ 2ρ + (2ρ +√

2G(R))RT (2.9)

and (2.10)

kΘ(u) − Θ(v)kHs≤√ 2

Z T 0

kf (ux)x− f (vx)xkHsdτ ≤√

2 ˜GT ku − vkX(T ), where ˜G(R) = ¯G(R, R). Differentiating with respect to t, we see that (2.11) ut(x, t) = (∂ttS(t)φ)(x) + (∂tS(t)ψ)(x) +

Z T 0

tΘ(t − τ )f (ux)xdτ.

Using the estimates in Lemmas 2.1 and 2.2, we have kΘ(u)tkHs ≤ 2kφkHs+ kψkHs−2+

Z T 0

kf (ux)xkHs−4

≤ 2(ρ + G(R)RT ) (2.12)

and

kΘ(u)t− Θ(v)tkHs ≤ Z T

0

kf (ux)x− f (vx)xkHs−4

≤ G(R)T ku − vkX(T ). (2.13)

From (2.9)-(2.13), we have

kΘ(u)kX(T )≤ 4ρ + (2ρ + 2√

3G(R)R)T, kΘ(u) − Θ(v)kX(T )≤ 3 ¯G(R)T ku − vkX(T ).

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Choosing R = 8ρ and fixing T so small enough such that T < min

ρ+

3G(R)R,4 ¯G(R)1  , (2.14)

therefore, Θ is a contraction map on BR(T ). It follows from the contrac- tion mapping theorem that problem (2.1) and (1.2) has a unique solution u ∈ BR(T ). Similar to that of [20], we can prove uniqueness and local Lipschitz dependence with respect to the initial data in the space BR(T ). Using unique- ness we can extend the result in the space C([0, T0]; Hs(R)) by a standard

technique. 

Proof of Theorem 1.2. It follows from (1.1) that d

dtE(t) = (utt, ut) + (ux, uxt)

+ (uxx, uxxt) + (uxtt, uxt) + (uxxtt, uxxt) + (f (ux), uxt)

= h(utt− uxxtt+ uxxxxtt− uxx+ uxxxx− f (ux)x, utiX∗X= 0, where (·, ·) denotes the inner product of L2-space and h·, ·iX∗Xmeans the usual duality of X and X with X = H2. Integrating the above equality with respect

to t, we have (1.9). Theorem 1.2 is proved. 

Next we study the existence of global solutions to the problem (1.1) and (1.2).

Proof of Theorem 1.3. One implication is obvious in view of Theorem 1.1. Let us prove that if

(2.15) lim

t∈[0,T0)

sup kux(·, t)kL = M < ∞, then T0= ∞. (1.1) can be written as follows:

utt+ u = P u + P f (ux)x, where P = (1 − ∂x2+ ∂x4)−1.

So, it follows from H¨older inequality that for t ∈ (0, T ), 1

2 d

dt(kuk2Hs+ kutk2Hs)

= ((I − ∂x2)s2utt, (I − ∂2x)s2ut) + ((I − ∂x2)s2u, (I − ∂2x)s2ut)

= ((I − ∂x2)sutt+ (I − ∂x2)su, ut)

= ((I − ∂x2)s(utt+ u), ut)

= ((I − ∂x2)s2(P u + P (∂xf (ux))), (I − ∂x2)s2ut)

≤ k(I − ∂x2)s2(P u + P (∂xf (ux)))kHskutkHs

≤ (k(I − ∂2x)s2P ukHs+ k(I − ∂2x)s2P (∂xf (ux))kHs)kutkHs. From Lemma 2.4, Lemma 2.6 and (3.1), we obtain

(k(I − ∂x2)s2P ukHs+ k(I − ∂x2)s2P (∂xf (ux))kHs) ≤ (G(R) + C)kukHs.

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It follows from the Cauchy inequality that 1

2 d

dt(kuk2Hs+ kutk2Hs) ≤ (G(R)2+ C)(kuk2Hs+ kutk2Hs), t ∈ (0, T ).

Then previous relation shows by Gronwall’ inequality that ku(t)k2Hs+kut(t)k2Hs

do not blow-up in finite time. Then T = ∞ by Theorem 1.1. The theorem is

proved. 

3. Existence and nonexistence of global solutions for f (u) = r|u|p In this section, we discuss the existence and nonexistence of global solution for the low initial energy case and the critical energy case when f (s) = r|u|p, r 6=

0. We start with the following elementary statement.

Lemma 3.1. The depth of potential well d = 2(p+1)p−1 (|r|Sp+1)p−12 , where S is the optimal Sobolev constant, i.e.,

(3.1) S= sup

u∈H2\{0}

kuxkp+1

(kuxk2+ kuxxk2)12.

Proof. From the definition of d, we obtain u ∈ N E, i.e., I(u) = 0, which yields kuxk2+ kuxxk2= −rkuxkp+1p+1 ≤ |r|Sp+1(kuxk2+ kuxxk2)p−12 (kuxk2+ kuxxk2).

From (3.1), we get

kuxk2+ kuxxk2≥ (|r|Sp+1)p−12 . (3.2)

On the other hand, from (1.11), (1.12), (3.2) and I(u) = 0, we obtain J (u) = (1

2 − 1

p + 1)(kuxk2+ kuxxk2)

≥ p − 1

2(p + 1)(|r|Sp+1)p−12 , (3.3)

which completes the proof. 

Lemma 3.2. Let φ, ψ ∈ H2, and u ∈ C1([0, T0); H2) is the unique solution of the Cauchy problem (1.1) and (1.2), where T0is the maximal existence time of u(t). Assume that E(0) < d, then for all t ∈ [0, T0),

(i) u(t) ∈ K1 and kux(t)k2+ kuxx(t)k2<2(p+1)p−1 d if φ ∈ K1; (ii) u(t) ∈ K2 and kux(t)k2+ kuxx(t)k2> 2(p+1)p−1 d if φ ∈ K2.

Proof. Since the proof of (i) and (ii) are similar, we only prove (ii). Let u(t) be any local weak solution of problem (1.1) and (1.2) with E(0) < d, φ ∈ K2and T0

be the maximum existence time of u(t). Then, it follows from Theorem 1.2 that E(u(t)) = E(0) < d. Thus, it suffices to show that I(u(t)) < 0 for 0 < t < T0. Arguing by contradiction, we suppose that there exists a t1∈ (0, T0) such that

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I(u(t1)) ≥ 0. From the continuity of I(u(t)) in time, there exists a t∈ (0, T0) such that I(u(t)) = 0. Then, from the definition of d, we get

d ≤ J (u(t)) ≤ E(u(t)) = E(0) < d,

which is a contradiction, that is kuxk2+kuxxk2< −rkuxkp+1p+1for any t ∈ [0, T0).

From (3.1), we arrive at 1

S2 ≤kuxk2+ kuxxk2

kuxk2p+1 <−rkuxkp+1p+1

kuxk2p+1 ≤ |r|kuxkp−1p+1. (3.4)

Then, from Lemma 3.1 and (3.4), yields d = p − 1

2(p + 1)|r|p−12 S−2S

4

p−1

< p − 1 2(p + 1)

kuxk2+ kuxxk2

kuxk2p+1 (kuxkp−1p+1)p−12

= p − 1

2(p + 1)(kuxk2+ kuxxk2), which means

kuxk2+ kuxxk2> 2(p + 1) p − 1 d.

In fact, for (i), from the definition of I(t) and E(t), we get E(t) = 1

2[kutk2+ kuxtk2+ kuxk2+ kuxxk2+ kuxxtk2] + r p + 1

Z

R

|ux|puxdx

= 1

2(kutk2+ kuxtk2+ kuxxtk2) + p − 1

2(p + 1)(kuxk2+ kuxxk2) + 1 p + 1I(t).

If I(u(t)) > 0, we obtain p − 1

2(p + 1)(kuxk2+ kuxxk2) < E(t) = E(0) < d.

So,

kuxk2+ kuxxk2< 2(p + 1) p − 1 d.

This completes the proof of Lemma 3.2. 

Lemma 3.3. Let φ, ψ ∈ H2, and u ∈ C1([0, T0); H2) is the unique solution of the Cauchy problem (1.1) and (1.2), where T0is the maximal existence time of u(t). Assume that E(0) = d and (φxx, ψxx) + (φx, ψx) + (φ, ψ) ≥ 0, then for all t ∈ [0, T0), u(t) ∈ K2 if u(0) ∈ K2.

Proof. If the result is false, there would exist a t0∈ (0, T0) such that I(t0) = 0 from the continuity of I(t). Hence we have J (u(t0)) ≥ d, which together with E(t0) = E(0) = d gives J (u(t0)) = d and

kut(t0)k + kuxt(t0)k + kuxxt(t0)k = 0.

(3.5)

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On the other hand, let

L(t) = ku(t)k2+ kux(t)k2+ kuxx(t)k2. (3.6)

Then

L0(t) = 2(u, ut) + 2(ux, uxt) + 2(uxx, uxxt), (3.7)

with L0(0) ≥ 0. From (1.1) and (1.8), we get

L00(t) = 2hutt− uxxtt+ uxxxxtt, uiX∗X+ 2kutk2+ 2kuxtk2+ 2kuxxtk2

= −2kuxk2−2kuxxk2−2r Z

R

|ux|puxdx + 2kutk2+ 2kuxtk2+ 2kuxxtk2

= 2kutk2+ 2kuxtk2+ 2kuxxtk2− 2I(u)

> 0, ∀t ∈ (0, t0).

(3.8)

Hence L0(t) is strictly increasing on [0, t0], and L0(t0) > 0 which contradicts

(3.5). The Lemma is proved. 

Lemma 3.4. Let φ ∈ Hs, ψ ∈ Hs, 2 ≤ s ≤ p + 1. Assume that E(0) < d, then when φ ∈ K1, problem (1.1) and (1.2) has a unique solution u ∈ C1([0, ∞); Hs) and u ∈ K1 for 0 ≤ t < ∞.

Proof. From Lemma 3.2, we can obtain

kux(t)k2+ kuxx(t)k2< 2(p + 1)d p − 1 .

It follows from above inequality and the Sobolev imbedding theorem that sup

t∈[0,T0)

kux(·, t)k2L < C2(p + 1)d p − 1 .

Therefore, problem (1.1) and (1.2) has a unique global solution u ∈ C1([0, ∞);

Hs) by Theorem 1.3. 

Proof of Theorem 1.4. First, it follows from Theorem 1.3 that problem (1.1) and (1.2) has a unique local solution u ∈ C1([0, T0), Hs), where T0 is the maximal existence time of u(t). Next we prove T0= ∞.

It follows from φ ∈ K1∪ ∂K1, E(0) ≤ d and the proof of Lemma 3.2 that kφxkp+1p+1≤ Sp+1(kφxk2+ kφxxk2)p+12

≤ Sp+1(kφxk2+ kφxxk2)(2(p + 1)d p − 1 )p−12

= |r|−1(kφxk2+ kφxxk2).

Therefore for any λ > 0, we obtain J (λu) = λ2

2 [kuxk2+ kuxxk2] +rλp+1 p + 1

Z

R

|ux|puxdx

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and

d

dλJ (λφ) = λ(kφxk2+ kφxxk2) + rλp Z

R

x|pφxdx

≥ λ(kφxk2+ kφxxk2) − |r|λpxkp+1p+1

≥ λ(1 − λ)(kφxk2+ kφxxk2) ≥ 0, ∀λ ∈ (0, 1).

Take a sequence {λm} such that 0 < λm < 1, m = 1, 2, . . . and λm → 1 as m → ∞. Let φm = λmφ, ψm = λmψ. Consider the problem (1.1) with the initial conditions

u(x, 0) = φm(x), ut(x, 0) = ψm(x).

(3.9) Then

mxk2+ kφmxxk2= λ2m[kφmxk2+ kφmxxk2] < 2(p + 1)d p − 1 and

Em(0) = 1

2[kψmk2+ kψmxk2+ kψmxxk2 + kφmxk2+ kφmxxk2] + r

p + 1 Z

R

mx|pφmxdx,

= 1

2[kψmk2+ kψmxk2+ kψmxxk2] + J (λmφ).

If ψ = 0 and φ = 0, then Em(0) = 0 < d. If ψ 6= 0 and φ 6= 0, then Em(0) < 1

2[kψmk2+ kψmxk2+ kψmxxk2] + J (φ) = E(0) ≤ d.

It follows from Lemma 3.4 that for each m, problem (1.1) and (3.9) has a unique global solution um∈ C2([0, ∞); Hs) and it satisfies

(umt, v) + (umxt, vx) + (umxxt, vxx) +

Z t 0

[(umx, vx) + (umxx, vxx) + (r|umx|p, vx)]dτ

= (ψm, v) + (ψmx, vx) + (ψmxx, vx), ∀v ∈ H2, t ∈ [0, ∞) (3.10)

and

1

2[kumtk2+ kumxtk2+ |umxxtk2] + J (um) = Em(0) < d, (3.11)

I(um) > 0, (3.12)

kumxk2+ kumxxk2< 2(p + 1) p − 1 d.

(3.13)

By using (3.11)-(3.13), we get

kumxkp+1p+1≤ Sp+1(kumxk2+ kumxxk2)p+12 ≤ |r|−1(kumxk2+ kumxxk2), J (um) ≥ 1

2[kumxk2+ kumxxk2] − |r|

p + 1 Z

R

|umx|p+1dx

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≥ p − 1

2(p + 1)[kumxk2+ kumxxk2] ≥ 0.

Because of the above inequalities, we have

kumtk2+ kumxtk2+ kumxxtk2≤ 2d, (3.14)

kumkp+1p+1≤ |r|−12(p + 1) p − 1 d.

(3.15)

It follows from (3.11), (3.12), (3.14) and (3.15) that there exist a ¯u ∈ ¯K1 and a subsequence {uk} such that as k → ∞, ukx→ ¯uxin L(0, ∞; H1) weakly star and a.e in R × [0, ∞), ukt→ ¯ut in L(0, ∞; L2) weakly star, |ukx|p→ | ¯ux|pin L(0, ∞; Lp+1p ) weakly star.

In equality (3.10), letting m = k → ∞, we obtain (¯ut, v) + (¯uxt, vx) + (¯uxxt, vxx) +

Z t 0

[(¯ux, vx) + ¯uxx, vxx) + (r|¯ux|p, vx)]dτ

= (ψ, v) + (ψx, vx) + (ψxx, vxx), ∀t ∈ [0, ∞),

for any v ∈ H2, which implies that ¯u satisfies (1.1). Furthermore, we can get

¯

u(x, 0) = φ(x), ¯ut(x, 0) = ψ(x).

Thus ¯u is a global solution of the Cauchy problem (1.1) and (1.2). From the uniqueness of the solution of problem (1.1) and (1.2), we get ¯u = u on R×[0, T0), and I(u) = I(¯u) ≥ 0. We obtain T0= ∞ and u ∈ C2([0, ∞); Hs). The theorem

is proved. 

Proof of Theorem 1.5. Theorem 1.1 gives the existence of a local weak solution u ∈ C1([0, T0); Hs) satisfying (1.9), where T0 is the maximal existence time of u. We prove T0< ∞. We assume that the result is not true, then T0= ∞. Let

L(t) = ku(t)k2+ kuxk2+ kuxx(t)k2, (3.16)

then

L0(t) = 2(u, ut) + 2(ux, uxt) + 2(uxx, uxxt).

Using the Schwartz inequality, we have

L0(t)2≤ 4L(t)[kutk2+ kuxtk2+ kuxxtk2].

(3.17)

Similarly to (3.8), from Lemmas 3.2 and 3.3, we obtain

(3.18)

L00(t) = (p + 3)[kutk2+ kuxtk2+ kuxxtk2] + (p − 1)[kuxk2+ kuxxk2]

− 2(p + 1)E(0)

> (p + 3)[kutk2+ kuxtk2+ kuxxtk2] + 2(p + 1)[d − E(0)]

≥ 0, ∀t ∈ (0, ∞).

It follows that

L0(t) > L0(0) + 2(p + 1)[d − E(0)]t, ∀t ∈ (0, ∞).

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This implies that there is t1 > 0 such that for any t ∈ [t1, ∞), L0(t) > 0.

Consequently, L(t) never vanishes on [t1, ∞). On the other hand, it follows from (3.16)-(3.18) that

L(t)L00(t) − (1 +p − 1

4 )L0(t)2> 0.

Set N (t) = (L(t))p−14 . Then we get N00(t) = −p − 1

4 (L(t))p+74 [L(t)L00(t) − (1 +p − 1

4 )L0(t)2] < 0, ∀t ∈ [t1, ∞) as well as N (t1) > 0 and N0(t1) < 0. Thus N (t) ≤ N (t1) + (t − t1)N0(t1).

So there is a T1∈ (t1, t1+(p−1)L4L(t10(t)1)) such that lim

t→T1

L(t) = ∞,

which contradicts T0= ∞. The theorem is proved.  4. Global existence for E(0) > 0

For the arbitrary initial energy E(0) > 0, using the technique of [4, 24], we derive a sufficient condition on the initial data such that the corresponding local solution of problem (1.1) and (1.2) exists globally.

Lemma 4.1. Let 2 ≤ s < p + 1, φ ∈ Hs∩ H01, ψ ∈ Hs and u(x, t) be the solution of the Cauchy problem (1.1) and (1.2). Assume that the initial data satisfy (1.14) and (1.16). Then, the map

{t → kuk2+ kuxk2+ kuxxk2} is strictly decreasing as long as u(x, t) ∈ YT.

Proof. From (3.6), (3.7) and (3.8), we have

L00(t) = 2kutk2+ 2kuxtk2+ 2kuxxtk2− 2I(t).

(4.1)

Furthermore, from u(t) ∈ YT, we get L00(t) < 0 for t ∈ [0, T ). Using (1.16), we obtain

2(φ, ψ) + 2(φx, ψx) + 2(φxx, ψxx) < 0,

which implies L0(0) < 0. So it is easy to see that L0(t) < L0(0) < 0, namely L0(t) < 0. Therefore, we get the result of the lemma.  Lemma 4.2. Let 2 ≤ s < p + 1, φ ∈ Hs∩ H01, ψ ∈ Hs and u(x, t) be the weak solution of the Cauchy problem (1.1) and (1.2) with maximal existence time interval [0, T ) such that u ∈ C1((0, T ), Hs∩ H01) and ut∈ C((0, T ), Hs), here T ≤ +∞. Assume that the initial data satisfy (1.14), (1.15) and (1.16), then u ∈ YT.

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Proof. Arguing by contradiction, we suppose that there exists a first time t2∈ [0, T ) such that

I(u(t2)) = kut(t2)k2+ kuxt(t2)k2+ kuxxt(t2)k2 (4.2)

and

I(u(t)) > kut(t)k2+ kuxt(t)k2+ kuxxt(t)k2, ∀t ∈ [0, t2).

From the (3.6)-(3.8) and Lemma 4.1, we obtain that L(t) and L0(t) are both strictly decreasing on the interval [0, t2). And by (1.16), for all t ∈ (0, t2), we get

L(u(t)) < kφk2+ kφxk2+ kφxxk2

< −2(φ, ψ) − 2(φx, ψx) − 2(φxx, ψxx) −2(p + 1) p + 3 E(0).

Moreover, from the continuity of kuk2+ kuxk2+ kuxxk2 in t, we obtain (4.3) L(u(t2)) < −2(φ, ψ) − 2(φx, ψx) − 2(φxx, ψxx) −2(p + 1)

p + 3 E(0).

On the other hand, by Theorem 1.2, (1.10) and (1.11), we get E(0) = E(t2) = 1

2(kut(t2)k2+ kuxt(t2)k2+ kuxxt(t2)k2) + J (u(t2))

= 1

2(kut(t2)k2+ kuxt(t2)k2+ kuxxt(t2)k2) + p − 1

2(p + 1)(kux(t2)k2+ kuxx(t2)k2) + 1

p + 1I(u(t2)).

Using (4.2), we obtain E(0) = (1

2 + 1

p + 1)(kut(t2)k2+ kuxt(t2)k2+ kuxxt(t2)k2) + (1

2 − 1

p + 1)(kux(t2)k2+ kuxx(t2)k2)

≥ p + 3

2(p + 1)(kut(t2)k2+ kuxt(t2)k2+ kuxxt(t2)k2).

(4.4)

Then from the following equalities

kut(t2)k2= kut(t2) + u(t2)k2− ku(t2)k2− 2(u(t2), ut(t2)), kuxt(t2)k2= kuxt(t2) + ux(t2)k2− kux(t2)k2− 2(ux(t2), uxt(t2)), kuxxt(t2)k2= kuxxt(t2) + uxx(t2)k2− kuxx(t2)k2− 2(uxx(t2), uxxt(t2)) and Lemma 4.1, we get

E(0) ≥ p + 3

2(p + 1)(kut(t2) + u(t2)k2+ kuxt(t2) + ux(t2)k2 + kuxxt(t2) + uxx(t2)k2)

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− p + 3

2(p + 1)(ku(t2)k2+ kux(t2)k2+ kuxx(t2)k2)

− 2 p + 3

2(p + 1)((u(t2), ut(t2)) + (ux(t2), uxt(t2)) + (uxx(t2), utxx(t2)))

≥ − p + 3

2(p + 1)(ku(t2)k2+ kux(t2)k2+ kuxx(t2)k2)

− 2 p + 3

2(p + 1)((φ(t2), ψ(t2)) + (φx(t2), ψx(t2)) + (φxx(t2), ψxx(t2))).

(4.5) So, (4.6)

L(t2) ≥ −2((φ(t2), ψ(t2))+(φx(t2), ψx(t2))+(φxx(t2), ψxx(t2)))−2(p + 1) p + 3 E(0).

It is obvious that (4.6) contradicts (4.3). This completes the proof.  Proof of Theorem 1.6. From Theorem 1.1, there exists a unique local solution of problem (1.1) and (1.2) defined on a maximal time interval [0, T ), T <

+∞. Let u(t) be the weak solution of problem (1.1) and (1.2) such that u ∈ C1((0, T ), H1∩ H01) and ut∈ C((0, T ), H1) with (1.14), (1.15) and (1.16).

Then from Lemma 4.2, we have u(x, t) ∈ YT, namely for t ∈ [0, T ), I(u(t)) > kut(t)k2+ kuxt(t)k2+ kuxxtk2.

(4.7)

Therefore from Theorem 1.2, (1.10), (1.11) and (4.7), we obtain E(0) = E(t) = 1

2(kut(t)k2+ kuxt(t)k2+ kuxxt(t)k2) + J (u)

= 1

2(kut(t)k2+ kuxt(t)k2+ kuxxt(t)k2) + p − 1

2(p + 1)(kux(t)k2+ kuxx(t)k2) + 1

p + 1I(u(t))

> p + 3

2(p + 1)(kut(t)k2+ kuxt(t)k2+ kuxxt(t)k2) + p − 1

2(p + 1)(kux(t)k2+ kuxx(t)k2).

From the Poincar´e inequality, that is kvk2 ≤ C0kvxk2, ∀v ∈ H01, where C0 is a positive constant, it follows that u(x, t) is bounded in C1((0, T ), H2∩ H01), ut(x, t) is bounded in C1((0, T ), H2). Hence from Theorem 1.1, it follows that T = ∞ and the solution of problem (1.1) and (1.2) exists globally.  Acknowledgments. The work is partially supported by the NSF-China grant- 11701068, the Fundamental Research Funds for the Central Universities grant- ZYGX2015J096.

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Ying Wang

School of Mathematical Sciences

University of Electronic Science and Technology of China Chengdu 611731, P. R. China

Email address: nadine1979@163.com

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